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Re: eigenvector
Posted:
Jun 16, 2013 4:38 AM


"Remus " <remusac@yahoo.com> wrote in message <kpighh$qj3$1@newscl01ah.mathworks.com>...
> > Hi guys, > I'm also looking at this problem. To answer you question John, yes I know that 1p is an Eigenvector of A. Let A be a matrix which has sum of all rows = 0, then 1p (where p=dim(a)) is an eigenvector of A. > For Example let A = [0 0 0; 1 1 0; 1 0 1]. the comand [V,D]=eig(A) returns the following eigenvectors: > V = [0 0 .5774; 1 0 .5774; 0 1 .5774], which is correct but as posted in the thread it is not normalized to 1.
It *is* normalized to 1
>> A = [0 0 0; 1 1 0; 1 0 1]
A =
0 0 0 1 1 0 1 0 1
>> [V,D]=eig(A)
V =
0 0 0.5774 0 1.0000 0.5774 1.0000 0 0.5774
D =
1 0 0 0 1 0 0 0 0
>> sqrt(sum(V.^2,1)) % compute l2norm of 3 eigen vectors
ans =
1 1 1
>>
% Bruno



